Can someone give an example of a function that is not Henstock-Kurzweil/gauge integrable?
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I am looking for an example of a function that is not Henstock-Kurzweil integrable. Can anybody help me?
real-analysis integration examples-counterexamples gauge-integral
edited Dec 3 '17 at 16:18
Martin Sleziak
43.7k6113260
asked Jun 5 '14 at 8:42
Elke
162
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show 3 more comments
up vote
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I am looking for an example of a function that is not Henstock-Kurzweil integrable. Can anybody help me?
real-analysis integration examples-counterexamples gauge-integral
edited Dec 3 '17 at 16:18
Martin Sleziak
43.7k6113260
asked Jun 5 '14 at 8:42
Elke
162
1
I do not know the answer, but I feel like a function that has something to do with a non Lebesgue measurable set would do. Since Lebesgue integral is "absolute convergent" and gauge integral is "conditional convergent"; the power of gauge integral is to integrate highly oscillating functions.
– Xiao
Jun 5 '14 at 9:02
How about $1/x$, say on $(-1,1)$ ?
– user8268
Jun 5 '14 at 9:33
$1/x$ is Riemann integrable, so it is also Henstock-Kurzweil integrable, right?
– Elke
Jun 5 '14 at 11:12
No, a function must be bounded to be Riemann integrable, but I believe your second statement is true.
– Ted Shifrin
Jun 5 '14 at 11:38
Of course, it would be easy to come up with an unbounded functions that the integral sum is positive infinity. The goal would be finding a bounded function on $[a,b]$ that is not Henstock-Kurzweil integrable.
– Xiao
Jun 6 '14 at 11:21
 |Â
show 3 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I am looking for an example of a function that is not Henstock-Kurzweil integrable. Can anybody help me?
real-analysis integration examples-counterexamples gauge-integral
edited Dec 3 '17 at 16:18
Martin Sleziak
43.7k6113260
asked Jun 5 '14 at 8:42
Elke
162
I am looking for an example of a function that is not Henstock-Kurzweil integrable. Can anybody help me?
real-analysis integration examples-counterexamples gauge-integral
real-analysis integration examples-counterexamples gauge-integral
edited Dec 3 '17 at 16:18
Martin Sleziak
43.7k6113260
asked Jun 5 '14 at 8:42
Elke
162
edited Dec 3 '17 at 16:18
Martin Sleziak
43.7k6113260
asked Jun 5 '14 at 8:42
Elke
162
edited Dec 3 '17 at 16:18
Martin Sleziak
43.7k6113260
edited Dec 3 '17 at 16:18
Martin Sleziak
43.7k6113260
edited Dec 3 '17 at 16:18
Martin Sleziak
43.7k6113260
43.7k6113260
asked Jun 5 '14 at 8:42
Elke
162
asked Jun 5 '14 at 8:42
Elke
162
asked Jun 5 '14 at 8:42
Elke
162
162
1
I do not know the answer, but I feel like a function that has something to do with a non Lebesgue measurable set would do. Since Lebesgue integral is "absolute convergent" and gauge integral is "conditional convergent"; the power of gauge integral is to integrate highly oscillating functions.
– Xiao
Jun 5 '14 at 9:02
How about $1/x$, say on $(-1,1)$ ?
– user8268
Jun 5 '14 at 9:33
$1/x$ is Riemann integrable, so it is also Henstock-Kurzweil integrable, right?
– Elke
Jun 5 '14 at 11:12
No, a function must be bounded to be Riemann integrable, but I believe your second statement is true.
– Ted Shifrin
Jun 5 '14 at 11:38
Of course, it would be easy to come up with an unbounded functions that the integral sum is positive infinity. The goal would be finding a bounded function on $[a,b]$ that is not Henstock-Kurzweil integrable.
– Xiao
Jun 6 '14 at 11:21
 |Â
show 3 more comments
1
I do not know the answer, but I feel like a function that has something to do with a non Lebesgue measurable set would do. Since Lebesgue integral is "absolute convergent" and gauge integral is "conditional convergent"; the power of gauge integral is to integrate highly oscillating functions.
– Xiao
Jun 5 '14 at 9:02
How about $1/x$, say on $(-1,1)$ ?
– user8268
Jun 5 '14 at 9:33
$1/x$ is Riemann integrable, so it is also Henstock-Kurzweil integrable, right?
– Elke
Jun 5 '14 at 11:12
No, a function must be bounded to be Riemann integrable, but I believe your second statement is true.
– Ted Shifrin
Jun 5 '14 at 11:38
Of course, it would be easy to come up with an unbounded functions that the integral sum is positive infinity. The goal would be finding a bounded function on $[a,b]$ that is not Henstock-Kurzweil integrable.
– Xiao
Jun 6 '14 at 11:21
1
1
I do not know the answer, but I feel like a function that has something to do with a non Lebesgue measurable set would do. Since Lebesgue integral is "absolute convergent" and gauge integral is "conditional convergent"; the power of gauge integral is to integrate highly oscillating functions.
– Xiao
Jun 5 '14 at 9:02
I do not know the answer, but I feel like a function that has something to do with a non Lebesgue measurable set would do. Since Lebesgue integral is "absolute convergent" and gauge integral is "conditional convergent"; the power of gauge integral is to integrate highly oscillating functions.
– Xiao
Jun 5 '14 at 9:02
How about $1/x$, say on $(-1,1)$ ?
– user8268
Jun 5 '14 at 9:33
How about $1/x$, say on $(-1,1)$ ?
– user8268
Jun 5 '14 at 9:33
$1/x$ is Riemann integrable, so it is also Henstock-Kurzweil integrable, right?
– Elke
Jun 5 '14 at 11:12
$1/x$ is Riemann integrable, so it is also Henstock-Kurzweil integrable, right?
– Elke
Jun 5 '14 at 11:12
No, a function must be bounded to be Riemann integrable, but I believe your second statement is true.
– Ted Shifrin
Jun 5 '14 at 11:38
No, a function must be bounded to be Riemann integrable, but I believe your second statement is true.
– Ted Shifrin
Jun 5 '14 at 11:38
Of course, it would be easy to come up with an unbounded functions that the integral sum is positive infinity. The goal would be finding a bounded function on $[a,b]$ that is not Henstock-Kurzweil integrable.
– Xiao
Jun 6 '14 at 11:21
Of course, it would be easy to come up with an unbounded functions that the integral sum is positive infinity. The goal would be finding a bounded function on $[a,b]$ that is not Henstock-Kurzweil integrable.
– Xiao
Jun 6 '14 at 11:21
 |Â
show 3 more comments
1 Answer
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Take any infinite series $a$, and let $f(x)$ be $a_lfloorxrfloor$ (where $lfloorxrfloor$ is the greatest integer that is not greater than $x$, that is $x$ rounded down to an integer). In general, $$ int_i^j f(x) ,mathrmdx = sum_n = i^j-1 a_n ,$$ whenever $i$ and $j$ are integers. As stated, this is a fine Riemann integral, but change $j$ to $infty$ and we have:
- $ int_i^infty f(x) ,mathrmdx = sum_n = i^infty a_n $ as a Lebesgue integral iff $a$ is absolutely convergent;
- $ int_i^infty f(x) ,mathrmdx = sum_n = i^infty a_n $ as an improper Riemann integral iff $a$ is convergent (possibly conditionally);
- $ int_i^infty f(x) ,mathrmdx = sum_n = i^infty a_n $ also as a Henstock–Kurzweil integral iff $a$ is convergent (possibly conditionally).
So any divergent series gives a Henstock–Kurzweil nonintegrable function. The harmonic series, for example, gives $f(x) = 1/lfloorxrfloor$, which is not integrable.
OK, so just plain $1/x$ is an even simpler nonintegrable function, but I think that if you keep in mind the relationship between Lebesgue vs Henstock and absolute vs conditional convergence, then it will seem more obvious to you that, while Henstock–Kurzweil makes more functions integrable, it does not make all functions integrable.
answered Apr 29 '17 at 6:25
Toby Bartels
479412
For some people, Henstock–Kurzweil integrals are only defined on bounded intervals. Then the spirit of this answer does not work, although 1/x still works, only now the issue is near 0 rather than near infinity. But I don't deserve credit for that answer; @user8268 already gave it in the comments to the question.
– Toby Bartels
May 2 '17 at 22:04
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Take any infinite series $a$, and let $f(x)$ be $a_lfloorxrfloor$ (where $lfloorxrfloor$ is the greatest integer that is not greater than $x$, that is $x$ rounded down to an integer). In general, $$ int_i^j f(x) ,mathrmdx = sum_n = i^j-1 a_n ,$$ whenever $i$ and $j$ are integers. As stated, this is a fine Riemann integral, but change $j$ to $infty$ and we have:
- $ int_i^infty f(x) ,mathrmdx = sum_n = i^infty a_n $ as a Lebesgue integral iff $a$ is absolutely convergent;
- $ int_i^infty f(x) ,mathrmdx = sum_n = i^infty a_n $ as an improper Riemann integral iff $a$ is convergent (possibly conditionally);
- $ int_i^infty f(x) ,mathrmdx = sum_n = i^infty a_n $ also as a Henstock–Kurzweil integral iff $a$ is convergent (possibly conditionally).
So any divergent series gives a Henstock–Kurzweil nonintegrable function. The harmonic series, for example, gives $f(x) = 1/lfloorxrfloor$, which is not integrable.
OK, so just plain $1/x$ is an even simpler nonintegrable function, but I think that if you keep in mind the relationship between Lebesgue vs Henstock and absolute vs conditional convergence, then it will seem more obvious to you that, while Henstock–Kurzweil makes more functions integrable, it does not make all functions integrable.
answered Apr 29 '17 at 6:25
Toby Bartels
479412
For some people, Henstock–Kurzweil integrals are only defined on bounded intervals. Then the spirit of this answer does not work, although 1/x still works, only now the issue is near 0 rather than near infinity. But I don't deserve credit for that answer; @user8268 already gave it in the comments to the question.
– Toby Bartels
May 2 '17 at 22:04
add a comment |Â
up vote
3
down vote
Take any infinite series $a$, and let $f(x)$ be $a_lfloorxrfloor$ (where $lfloorxrfloor$ is the greatest integer that is not greater than $x$, that is $x$ rounded down to an integer). In general, $$ int_i^j f(x) ,mathrmdx = sum_n = i^j-1 a_n ,$$ whenever $i$ and $j$ are integers. As stated, this is a fine Riemann integral, but change $j$ to $infty$ and we have:
- $ int_i^infty f(x) ,mathrmdx = sum_n = i^infty a_n $ as a Lebesgue integral iff $a$ is absolutely convergent;
- $ int_i^infty f(x) ,mathrmdx = sum_n = i^infty a_n $ as an improper Riemann integral iff $a$ is convergent (possibly conditionally);
- $ int_i^infty f(x) ,mathrmdx = sum_n = i^infty a_n $ also as a Henstock–Kurzweil integral iff $a$ is convergent (possibly conditionally).
So any divergent series gives a Henstock–Kurzweil nonintegrable function. The harmonic series, for example, gives $f(x) = 1/lfloorxrfloor$, which is not integrable.
OK, so just plain $1/x$ is an even simpler nonintegrable function, but I think that if you keep in mind the relationship between Lebesgue vs Henstock and absolute vs conditional convergence, then it will seem more obvious to you that, while Henstock–Kurzweil makes more functions integrable, it does not make all functions integrable.
answered Apr 29 '17 at 6:25
Toby Bartels
479412
For some people, Henstock–Kurzweil integrals are only defined on bounded intervals. Then the spirit of this answer does not work, although 1/x still works, only now the issue is near 0 rather than near infinity. But I don't deserve credit for that answer; @user8268 already gave it in the comments to the question.
– Toby Bartels
May 2 '17 at 22:04
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Take any infinite series $a$, and let $f(x)$ be $a_lfloorxrfloor$ (where $lfloorxrfloor$ is the greatest integer that is not greater than $x$, that is $x$ rounded down to an integer). In general, $$ int_i^j f(x) ,mathrmdx = sum_n = i^j-1 a_n ,$$ whenever $i$ and $j$ are integers. As stated, this is a fine Riemann integral, but change $j$ to $infty$ and we have:
- $ int_i^infty f(x) ,mathrmdx = sum_n = i^infty a_n $ as a Lebesgue integral iff $a$ is absolutely convergent;
- $ int_i^infty f(x) ,mathrmdx = sum_n = i^infty a_n $ as an improper Riemann integral iff $a$ is convergent (possibly conditionally);
- $ int_i^infty f(x) ,mathrmdx = sum_n = i^infty a_n $ also as a Henstock–Kurzweil integral iff $a$ is convergent (possibly conditionally).
So any divergent series gives a Henstock–Kurzweil nonintegrable function. The harmonic series, for example, gives $f(x) = 1/lfloorxrfloor$, which is not integrable.
OK, so just plain $1/x$ is an even simpler nonintegrable function, but I think that if you keep in mind the relationship between Lebesgue vs Henstock and absolute vs conditional convergence, then it will seem more obvious to you that, while Henstock–Kurzweil makes more functions integrable, it does not make all functions integrable.
answered Apr 29 '17 at 6:25
Toby Bartels
479412
Take any infinite series $a$, and let $f(x)$ be $a_lfloorxrfloor$ (where $lfloorxrfloor$ is the greatest integer that is not greater than $x$, that is $x$ rounded down to an integer). In general, $$ int_i^j f(x) ,mathrmdx = sum_n = i^j-1 a_n ,$$ whenever $i$ and $j$ are integers. As stated, this is a fine Riemann integral, but change $j$ to $infty$ and we have:
- $ int_i^infty f(x) ,mathrmdx = sum_n = i^infty a_n $ as a Lebesgue integral iff $a$ is absolutely convergent;
- $ int_i^infty f(x) ,mathrmdx = sum_n = i^infty a_n $ as an improper Riemann integral iff $a$ is convergent (possibly conditionally);
- $ int_i^infty f(x) ,mathrmdx = sum_n = i^infty a_n $ also as a Henstock–Kurzweil integral iff $a$ is convergent (possibly conditionally).
So any divergent series gives a Henstock–Kurzweil nonintegrable function. The harmonic series, for example, gives $f(x) = 1/lfloorxrfloor$, which is not integrable.
OK, so just plain $1/x$ is an even simpler nonintegrable function, but I think that if you keep in mind the relationship between Lebesgue vs Henstock and absolute vs conditional convergence, then it will seem more obvious to you that, while Henstock–Kurzweil makes more functions integrable, it does not make all functions integrable.
answered Apr 29 '17 at 6:25
Toby Bartels
479412
edited Aug 31 at 7:01
answered Apr 29 '17 at 6:25
Toby Bartels
479412
answered Apr 29 '17 at 6:25
Toby Bartels
479412
answered Apr 29 '17 at 6:25
Toby Bartels
479412
479412
For some people, Henstock–Kurzweil integrals are only defined on bounded intervals. Then the spirit of this answer does not work, although 1/x still works, only now the issue is near 0 rather than near infinity. But I don't deserve credit for that answer; @user8268 already gave it in the comments to the question.
– Toby Bartels
May 2 '17 at 22:04
add a comment |Â
For some people, Henstock–Kurzweil integrals are only defined on bounded intervals. Then the spirit of this answer does not work, although 1/x still works, only now the issue is near 0 rather than near infinity. But I don't deserve credit for that answer; @user8268 already gave it in the comments to the question.
– Toby Bartels
May 2 '17 at 22:04
For some people, Henstock–Kurzweil integrals are only defined on bounded intervals. Then the spirit of this answer does not work, although 1/x still works, only now the issue is near 0 rather than near infinity. But I don't deserve credit for that answer; @user8268 already gave it in the comments to the question.
– Toby Bartels
May 2 '17 at 22:04
For some people, Henstock–Kurzweil integrals are only defined on bounded intervals. Then the spirit of this answer does not work, although 1/x still works, only now the issue is near 0 rather than near infinity. But I don't deserve credit for that answer; @user8268 already gave it in the comments to the question.
– Toby Bartels
May 2 '17 at 22:04
add a comment |Â
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1
I do not know the answer, but I feel like a function that has something to do with a non Lebesgue measurable set would do. Since Lebesgue integral is "absolute convergent" and gauge integral is "conditional convergent"; the power of gauge integral is to integrate highly oscillating functions.
– Xiao
Jun 5 '14 at 9:02
How about $1/x$, say on $(-1,1)$ ?
– user8268
Jun 5 '14 at 9:33
$1/x$ is Riemann integrable, so it is also Henstock-Kurzweil integrable, right?
– Elke
Jun 5 '14 at 11:12
No, a function must be bounded to be Riemann integrable, but I believe your second statement is true.
– Ted Shifrin
Jun 5 '14 at 11:38
Of course, it would be easy to come up with an unbounded functions that the integral sum is positive infinity. The goal would be finding a bounded function on $[a,b]$ that is not Henstock-Kurzweil integrable.
– Xiao
Jun 6 '14 at 11:21